Question: Simplify the following expression and state the condition under which the simplification is valid. $z = \dfrac{k^2 - 81}{k + 9}$
First factor the polynomial in the numerator. The numerator is in the form ${a^2} - {b^2}$ , which is a difference of two squares so we can factor it as $({a} + {b})({a} - {b})$ $ a = k$ $ b = \sqrt{81} = 9$ So we can rewrite the expression as: $z = \dfrac{({k} + {9})({k} {-9})} {k + 9} $ We can divide the numerator and denominator by $(k + 9)$ on condition that $k \neq -9$ Therefore $z = k - 9; k \neq -9$